Bài 5:

a: ĐKXĐ: x>0; x<>9

b: \(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{3}{\sqrt{x}-3}\)

\(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\)

\(=\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\)

\(=\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}=\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

Ta có: \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\frac{-3}{\sqrt{x}-3}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}=\frac{-3}{\sqrt{x}-3}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)

c: C<-1

=>C+1<0

=>\(\frac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}<0\)

=>\(\frac{-\sqrt{x}+4}{2\sqrt{x}+4}<0\)

=>\(-\sqrt{x}+4<0\)

=>\(-\sqrt{x}<-4\)

=>\(\sqrt{x}>4\)

=>x>16

Bài 4:

a: \(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

Ta có: \(P=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

\(=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

b: \(x=\frac{2}{2+\sqrt3}=\frac{2\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=4-2\sqrt3\)

Thay \(x=4-2\sqrt3\) vào P, ta được:

\(P=\frac{\left(\sqrt{4-2\sqrt3}+1\right)^2}{\sqrt{4-2\sqrt3}}=\frac{\left(\sqrt{\left(\sqrt3-1\right)^2}+1\right)^2}{\sqrt{\left(\sqrt3-1\right)^2}}\)

\(=\frac{\left(\sqrt3-1+1\right)^2}{\sqrt3-1}=\frac{3}{\sqrt3-1}=\frac{3\left(\sqrt3+1\right)}{2}=\frac{3\sqrt3+3}{2}\)

c: \(P\cdot\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)

=>\(\left(\sqrt{x}+1\right)^2=6\sqrt{x}-3-\sqrt{x-4}\)

=>\(x-4\sqrt{x}+4=-\sqrt{x-4}\)

=>\(\left(\sqrt{x}-2\right)^2=-\sqrt{x-4}\)

=>\(\begin{cases}\sqrt{x}-2=0\\ x-4=0\end{cases}\Rightarrow x=4\) (nhận)

Bài 3:

a: \(D=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b: D<1

=>D-1<0

=>\(\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}<0\)

=>\(\frac{4}{\sqrt{x}-3}<0\)

=>\(\sqrt{x}-3<0\)

=>\(\sqrt{x}<3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: 0<=x<9 và x<>4

c: D nguyên khi \(\sqrt{x}+1\) ⋮\(\sqrt{x}-3\)

=>\(\sqrt{x}-3+4\) ⋮\(\sqrt{x}-3\)

=>4⋮\(\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\lbrace1;-1;2;-2;4;-4\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace4;2;5;1;7;-1\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace1;2;4;5;7\right\rbrace\)

=>x∈{1;4;16;25;49}

kết hợp ĐKXĐ, ta được: x∈{1;16;25;49}

Bài 2:

a: \(\frac{2+\sqrt{a}}{2-\sqrt{a}}-\frac{2-\sqrt{a}}{2+\sqrt{a}}-\frac{4a}{a-4}\)

\(=\frac{-\left(\sqrt{a}+2\right)}{\sqrt{a}-2}+\frac{\sqrt{a}-2}{\sqrt...